IA Financial Group Aptitude Interview Questions & Answers

  1. Question 1. If A Number Cube Is Rolled Once And Coin Is Tossed Once. What Is The Probability That The Coin Will Show Tails And Composite Number On The Cube?

    Answer :

    Numbers on cube (1,2,3,4,5,6)

    Composite number on dice are 4 and 6.

    So, Probability that the coin will show tails and composite number on the cube = (1/2) *(2/6) = 1/6

    Note: Composite number is a positive integer that can be formed by multiplying together two smaller positive integers. Equivalently, it is a positive integer that has at least one divisor other than 1 and itself.

  2. Question 2. The Time A Passenger Train Takes To Cross Another Freight Train Is Twice When The Passenger Train Crosses The Freight Train Running In Opposite Directions. What Is The Ratio Of Their Speeds?

    Answer :

    Lets assume speed of freight train = X Km/hr

    speed of passenger train = Y Km/hr.

    and sum of their length = D Km.

    When trains are running in same direction:

    so Time = D/(X-Y) ———- (1)

    When trains are running in opposite direction:

    So, Time = D/(X+Y) ——— (2)

    By eq. (1) & (2)

    [D/(X-Y)] = 2 * [D/(X+Y)]

    => 2X-2Y = X+Y

    => X = 3Y 

    => X:Y = 3:1.

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  4. Question 3. The Banker’s Gain On A Bill Due 2 Year Hence At 10% Per Annum Is Rs. 8. The True Discount Is?

    Answer :

    True Discount = (B.G*100)/(Rate*Time) 

    T.D = (8*100)/(10*2)

    = 8*5 = Rs. 40.

  5. Question 4. If X(a) = A^2+2a-1, Then X(8)-x(5) =?

    Answer :

    x(8) = 64 + 16 – 1 = 79

    x(5) = 25 + 10 – 1 = 34

    So, x(8)-x(5) = 79 – 34 = 45.

  6. Question 5. If A Certain Computer Is Capable Of Printing 4900 Monthly Credit Card Bills Per Hour, While A New Model Is Capable Of Printing At A Rate Of 6600 Per Hour, The Old Model Will Take Approximately How Much Longer Than The New Model To Print 10000 Bills?

    Answer :

    Old model is capable of printing at a rate of 4900 per hour

    New model is capable of printing at a rate of 6600 per hour

    Old model time taken to print 10000 cards = 10000/4900 = 100/49

    New model time taken to print 10000 cards = 10000/6600 = 100/66

    Old model – new model: 100/49 – 100/66 = 1700 /(49*66) = 850/(49 * 33) = 0.525 hrs => 31 mins

    Therefore, the old model will take approximately 31 mins longer than the new model to print 10000 bills.

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  8. Question 6. To Finish A Piece Of Work, P Takes Five Times As Q And Six Times As R. Working Together, They Can Finish The Work In 4 Days. Q Can Do The Work Alone In?

    Answer :

    Lets assume Q take x days to finish the work

    so P take time = 5x, and R take time R=5x/6 

    Given, They took 4 day to finish the work.

    1/5x+1/x+6/5x= 1/4 => 12/5x = 1/4 

    x=48/5

    So, Q can do work alone in 48/5 days.

  9. Question 7. Total Of The Ages Of A, B Ahd C At Present Is 90 Years. Ten Years Ago, The Ratio Of Their Ages Was 1: 2: 3. What Is The Age Of B At Present?

    Answer :

    Let their ages 10 years ago is x, 2x and 3x years.

    (x + 10) + (2x + 10) + (3x + 10) = 90 hence x= 10

    B’s present age = (2x + 10) =30 years.

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  11. Question 8. The Sum Of A Number And Its Square Is 1406. What Is The Number?

    Answer :

    Lets assume the no. = x

    so x+x^2=1406

    by putting values from options

    37+37^2

    =37(37+1)

    =37*38 =1406.

  12. Question 9. Cost Of An Item Is X. It’s Value Increases By P% And Decreases By P% Now The New Value Is 1 Rupee, What Is The Actual Value ?

    Answer :

    Given cost of an item = Rs. X

    Item cost by increasing p% will be = x(100+p)/100

    After p% decrement, value = {x+(xp/100)}-[{x+(xp/100)}p/100] 

    = [x+(xp/100)] [1-(p/100)]

    = x[1+(p/100)][1-(p/100)]

    Givne new value is 1

    x[1-((p^2)/10000)] = 1

    x = 1/[1-((p^2)/10000)] = 10000/[10000-(p^2)].

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  14. Question 10. How Many Matches Would Have To Be Played To Decide A Winner In A Knockout Tournament Consisting Of Eight Teams?

    Answer :

    1 match for 2 team, first 8 team = 4 match

    4 team – 2 match

    2 team – 1 match

    So, Total = 4+2+1=7 match.

  15. Question 11. A Man Traveled From The Village To The Post-office At The Rate Of 25 Kmph And Walked Back At The Rate Of 4 Kmph. If The Whole Journey Took 5 Hours 48 Minutes, Find The Distance Of The Post-office From The Village?

    Answer :

    Average speed = (2*a*b)/(a + b) here a = 25 b = 4

    Average speed = 2*25 * 4/(25 + 4) = 200/29 kmph.

    Distance covered in 5 hours 48 minutes = Speed * time

    Distance = (200/29)*(29/5) = 40 kms

    Distance covered in 5 hours 48 minutes = 40 kms

    Distance of the post office from the village = (40/2) =20 km.

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  17. Question 12. Three Friends Gerard, Rooney And Ronaldson Work Together To Dig A Hole. Gerard Alone Can Complete The Work In 10 Days, Rooney In 8 Days And Together All Three Can Complete It In 4 Days. They Earn A Total Of Rs. 1,200. Find The Share Of Rooney If The Money That They Receive Is Proportional To The Work That They Do?

    Answer :

    Ronaldson can complete work in 40 days ( 1/4-1/10-1/8)

    So, ratio of work done = 1/10 :1/8 : 1/40 = 4 : 5 : 1

    So, amount payed rooney = 1200 * (5 / (5+4+1) ) = 1200 * 1/2 = 600

    So, Rooney share = 1200/2 = 600.

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  19. Question 13. The Product Of Two Positive Numbers Is P. If Each Of The Numbers Is Increased By 2, The New Product Is How Much Greater Twice The Sum Of The Two Original Numbers?

    Answer :

    Lets assume number are x,y 

    So, there product = xy

    Given number increased by 2 

    so, now number are x+2,y+2.

    Given product is twice the sum of two original No. 

    => product = (x+2)(y+2) = 2(x+y) 

    => xy+2x+2y+4 =2(x+y)

    => xy+4

    so it increase by 4.

  20. Question 14. The Price Of A Jewel, Passing Through Three Hands, Rises On The Whole By 65%. If The First And Second Sellers Earned 20% And 25% Profit Respectively, Find The Percentage Profit Earned By The Third Seller?

    Answer :

    Lets assume the original price of the jewel = Rs. P 

    and profit earned by the third seller= x%.

    Then, ( 100 +x )% of 125% of 120% of P = 165% of p

    => [ ( 100+x )/100*125/100*120/100* P ] = [ 165/100* P ]

    => ( 100 + X ) = (165*100*100)/(125*120) = 110,

    => X = 10%.

  21. Question 15. In A Mixture Of Petrol And Kerosene, Petrol Is Only 99 Litres. If This Same Quantity Of Petrol Would Be Presented In Another Mixture Of Petrol And Kerosene Where Total Volume Would Be 198 Litres Less Than The Actual Mixture Then The Concentration Of Petrol In The Present Mixture Would Have Been 13.33% Point Less Than That. What
    Is The Concentration Of Petrol In Actual Mixture?

    Answer :

    Lets assume amount of petrol = x

    So, (x*100)/198=13.33

    x=26.3934 

    Now same amount of petrol is in 99 liters also.

    so the percentage in actual mixture is

    (26.3934*100)/99 =26.66%.

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  23. Question 16. In A Election Between Two Candidate 10% Of Votes Were Invalid Winner Got 70% Of Valid Votes And Was Elected By A Majority Of 900 Votes. Find The Total Number Of Votes?

    Answer :

    Lets assume votes = x 

    valid votes are 0.9x,

    Given 70% of 0.9x is 900 more than 30% of 0.9x,

    => 0.9*0.7-900=0.3*0.9x 

    => x=2500.

  24. Question 17. Log Base 12 144p=3 Then P= ?

    Answer :

    log base 12 144p=3

    => 144p=12^3

    => 144p=12^2 *12

    => 144p=144*12

    => p=12.

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  26. Question 18. 3 Men And 8 Women Complete A Task In Same Time As 6 Men And 2 Women Do. How Much Fraction Of Work Will Be Finished In Same Time If 4 Men And 5 Women Will Do That Task?

    Answer :

    3M + 8W = 6M + 2W ———(1)

    solving this we get

    M = 2W

    Therefore 3M + 8W = 14W

    4M + 5W =13W

    =>fraction of work is 13/14.

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  28. Question 19. The Sum Of Ages Of Suresh And His Father Is 78 Years.6 Years Hence Father’s Age Will Be Twice Of Suresh’s Age. Find Present Age Of Father?

    Answer :

    Lets assume suresh age = x years.

    So, father age = (78-x) years. 

    Condition : 6 years hence

    2* (x + 6) = (78 – x + 6)

    => 2x + 12 = 84 – x 

    => 3x = 72

    x = 24 years.

    Father age = 78 – 24 = 54 years.

  29. Question 20. In This Question A^b Means A Raised To The Power Of B Start With The Integers From 1 To 10^2012. Replace Each Of Them By The Sum Of Its Digits To Get A String Of 10^2012 Numbers. Keep Doing This Until You Get 10^2012 Single Digit Numbers. Let M Be The Number Of 1’s And N Be The Number Of 2’s. Then M – N?

    Answer :

    We divide all the numbers into groups of 9. (1 to 9), (10 to 18)…..

    Now when we divide each term in the first group by 9, we get 1, 2, 3 ..0 as remainders. Now digit sum is nothing but finding remainder when a number is divided by 9. 

    So the last term, 10^2012 gives remainder 1 when divided by 9. So there is one “1” extra than 2’s.

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  31. Question 21. Two Trains A And B Start Simultaneously From Stations X And Y Towards Each Other Respectively. After Meeting At A Point Between X And Y,train A Reaches Station Y In 9 Hours And Train B Reaches Station X In 4 Hours From The Time They Have Met Each Other. If The Speed Of Traina Is 36 Km/hr, What Is The Speed Of Train B?

    Answer :

    Naming the trains as A and B. Then,

    (A’s speed) : (B’s speed) = (b : a)^.5 

    where b and a is the time taken by trains B and A, to reach their destination after they meet.

    hence,

    36/B’s speed = (4: 9)^.5

    => B’s speed = 54km/hr.

  32. Question 22. Which Of The Following Values Of ‘n’ Satisfies The In-equality N2 – 24n + 143 < 0?

    Answer :

    n^2 – 24n + 143 < 0

    => (n-11)(n-13) < 0

    => 11 < n < 13.

  33. Question 23. A Car Starts From Rest And Accelerates Uniform Upto Speed 90 Km/hrs In 5 Sec. Total Distance Covered By Car In This Time Will Be?

    Answer :

    Speed of car at last instant of given interval 

    = 90*5/ 18 m/s = 25 m/s

    so average speed = 0+25 / 2

    Therefore distance = 5 * 25/2 = 125/2 = 62.5 m.

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  35. Question 24. If A Man Reduces The Selling Price Of A Fan From 400 To 380 His Loss Increases By 20% .what Is The Cost Price Of Fan ?

    Answer :

    Lets assume the cost price = x.

    So, initial loss = (x – 400).

    Given, 20% ( x – 400) = 20

    => (x – 400) = 100

    => x = 500.

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  37. Question 25. A Piece Of Ribbon 4 Yards Long Is Used To Make Bows Requiring 15 Inches Of Ribbon For Each. What Is The Maximum Number Of Bows That Can Be Made?

    Answer :

    1 yard = 3 feet = 3*12 = 36 inches

    4 yard = 4*36 = 144 inches

    The maximum number of bows will be 4 yards divided by 15 inches = 144/15 = 9.6.

  38. Question 26. Q Is As Much Younger Than R As He Is Older Than T. If The Sum Of The Ages Of R And T Is 50 Years, What Is Definitely The Difference Between R And Q’s Age?

    Answer :

    As per question R > Q > T 

    So, R – Q = Q – T and R + T = 50 

    By solving above 2 condition.

    => Q = 25

    As the difference between R & Q and Q & T is same

    So it is 25 years.

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  40. Question 27. In An Exam, Ajith, Sachu, Karna, Saheep And Ramesh Scored An Average Of 39 Marks. Saheep Scored 7 Marks More Than Ramesh. Ramesh Scored 9 Fewer Than Ajith. Sachu Scored As Many As Saheep And Ramesh Scored. Sachu And Karna Scored 110 Marks Between Them. If Ajith Scores 32 Marks Then How Many Marks Did Karna Score?

    Answer :

    Lets assume marks of Ajith, Sachu, Karna, Saheep and Ramesh respectively u, v, w, x, y

    So, as per the question:

    z+7=x —(i)

    u- 9=z —(ii)

    x+ z=v —(iii)

    v+w=110 —(iv)

    Given, u=32 —- (v)

    By solving eq. (i), (ii), (iii), (iv) and (v)

    w=57.

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  42. Question 28. Silu And Meenu Were Walking On The Road. Silu Said?

    Answer :

    Given that Meenu weight 29 Kgs plus half of her own weight.

    It means that 29 Kgs is the other half. So she weighs 58 Kgs.

    Solving mathematically, let’s assume that her weight is A Kgs.

    A = 29 + A/2

    2*A = 58 + A

    A = 58 Kgs.

  43. Question 29. A New Apartment Complex Purchased 60 Toilets And 20 Shower Heads. If The Price Of A Toilet Is Three Times The Price Of A Shower Head, What Percent Of The Total Cost Was The Cost Of All The Shower Heads?

    Answer :

    Lets assume the cost of shower head = x. 

    So,cost of the toilet = 3x.

    Total cost of 60 toilets and 20 shower heads = 60 * 3x + 20* x = 200x

    So shower heads cost as a percentage of total cost = (200x/20x)% = 10%.

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  45. Question 30. Two Guys Work At Some Speed. After Some Time One Guy Realises He Has Done Only Half Of The Other Guy Completed Which Is Equal To Half Of What Is Left So How Much Faster Than The Other Is This Guy Supposed To Do To Finish With The First?

    Answer :

    If x is the part of task this is completed then,

    1st has done work = (1 – x)/4 

    and 2nd has done work = (1- x)/2

    So 2nd will have to increase his speed by 2 times.